UPSC 2023 Maths Optional Paper 1 Q8a — Step-by-Step Solution

15 marks · Section B

Laplace transform applied to IVP for second-order linear ODE with constant coefficients · ODEs · asked 10× in 13 yrs · Read the full method →

Question

Solve the following initial value problem by using Laplace transform technique:

\frac{d^2 y}{dt^2}-4\frac{dy}{dt}+3y(t)=f(t),\quad y(0)=1,\;y'(0)=0,

where $f(t)$ is a given function of $t$ .

Technique

Take Laplace transform of the ODE; partial fractions on the IC-driven part; convolution theorem to invert the $F(s)$ -driven part. The two pieces add to give the full solution.

Solution

Step 1 — Take Laplace transforms.

Let $Y(s)=L\{y(t)\}$ and $F(s)=L\{f(t)\}$ . Standard transforms: $L\{y''\}=s^2Y-sy(0)-y'(0)=s^2Y-s.$ $L\{y'\}=sY-y(0)=sY-1.$

Apply to the ODE:

s^2Y-s-4(sY-1)+3Y=F(s).

Y(s^2-4s+3)=F(s)+s-4.

Factor $s^2-4s+3=(s-1)(s-3)$ :

Y(s)=\frac{F(s)+s-4}{(s-1)(s-3)}.

Step 2 — Inverse-transform the IC-driven part.

$\dfrac{s-4}{(s-1)(s-3)}=\dfrac{A}{s-1}+\dfrac{B}{s-3}$ . Solve: $A(s-3)+B(s-1)=s-4$ . At $s=1$ : $-2A=-3\Rightarrow A=\tfrac{3}{2}$ . At $s=3$ : $2B=-1\Rightarrow B=-\tfrac{1}{2}$ .

$L^{-1}\Bigl\{\dfrac{s-4}{(s-1)(s-3)}\Bigr\}=\tfrac{3}{2}e^t-\tfrac{1}{2}e^{3t}.$

Step 3 — Inverse-transform the $F$ -driven part.

$\dfrac{1}{(s-1)(s-3)}=\dfrac{1}{2(s-3)}-\dfrac{1}{2(s-1)}$ (partial fractions).

L^{-1}\Bigl\{\frac{F(s)}{(s-1)(s-3)}\Bigr\}=\frac{1}{2}\bigl[L^{-1}\{F(s)/(s-3)\}-L^{-1}\{F(s)/(s-1)\}\bigr].

By the convolution theorem $L^{-1}\{F\cdot G\}=f*g$ : $L^{-1}\{F(s)/(s-3)\}=L^{-1}\{F(s)\}*L^{-1}\{1/(s-3)\}=f(t)*e^{3t}=\int_0^t f(\tau)e^{3(t-\tau)}d\tau.$ $L^{-1}\{F(s)/(s-1)\}=f(t)*e^t=\int_0^t f(\tau)e^{t-\tau}d\tau.$

Therefore

L^{-1}\Bigl\{\frac{F(s)}{(s-1)(s-3)}\Bigr\}=\frac{1}{2}\int_0^t f(\tau)\bigl[e^{3(t-\tau)}-e^{t-\tau}\bigr]d\tau.

Step 4 — Combine.

Answer

\boxed{\;y(t)=\frac{3}{2}e^t-\frac{1}{2}e^{3t}+\frac{1}{2}\int_0^t f(\tau)\bigl[e^{3(t-\tau)}-e^{t-\tau}\bigr]d\tau.\;}